How to Calculate Heat of Combustion and Heat Transfer in a Furnace?

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The discussion focuses on calculating the standard heat of combustion for a natural gas fuel mixture and the heat transfer in a furnace. Participants clarify that the exiting temperature of the combustion products does not equate to the reaction temperature, as the heat generated raises the temperature of the gases. The correct approach for calculating the heat of combustion involves using Hess's Law and considering the contributions from methane and ethane while noting that nitrogen does not react. Additionally, when calculating heat transfer, it is important to maintain consistency in the basis used for calculations, switching from one mole to a larger quantity if necessary. Overall, the calculations hinge on understanding the thermodynamic principles involved in combustion and heat transfer.
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Homework Statement


A natural-gas fuel contains 85 mol-% methane, 10 mol-% ethane, and 5 mol-% nitrogen.
a) What is the standard heat of combustion (kJ mol-1) of the fuel at 25°C with H2O(g) as a
product?
b) The fuel is supplied to a furnace with 50% excess air, both entering at 25°C. The products
leave at 600°C. If combustion is complete and if no side reactions occur, how much heat
(kJ mol-1 of fuel) is transferred in the furnace?


Homework Equations





The Attempt at a Solution


For this problem, I am unsure if in general, the exiting temperature is the reaction temperature. I mean, initially the gases are at 25 C, but when they exit they are 600C. That doesn't mean it reacted at 600 C, the result of the reaction just happened to be that temperature increase, right?
 

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Maylis said:

Homework Statement


A natural-gas fuel contains 85 mol-% methane, 10 mol-% ethane, and 5 mol-% nitrogen.
a) What is the standard heat of combustion (kJ mol-1) of the fuel at 25°C with H2O(g) as a
product?
b) The fuel is supplied to a furnace with 50% excess air, both entering at 25°C. The products
leave at 600°C. If combustion is complete and if no side reactions occur, how much heat
(kJ mol-1 of fuel) is transferred in the furnace?


Homework Equations





The Attempt at a Solution


For this problem, I am unsure if in general, the exiting temperature is the reaction temperature. I mean, initially the gases are at 25 C, but when they exit they are 600C. That doesn't mean it reacted at 600 C, the result of the reaction just happened to be that temperature increase, right?

Yes. You have the right idea. I didn't check every last detail of what you did, but you certainly knew to use Hess's Law. The heat given off by the reaction at 25C minus the heat loss from the reactor is enough to raise the temperature of the products from 25C to 600C.

chet
 
Does nitrogen combust with oxygen?

Also, is my part (a) correct? A friend of mine is trying it my doing .85ΔHf,methane + .10ΔHf,ethane + .05ΔHf,N2 - ΔHf, CO2 - ΔHf,H2O(g)
 
Last edited:
Maylis said:
Does nitrogen combust with oxygen?

Also, is my part (a) correct? A friend of mine is trying it my doing .85ΔHf,methane + .10ΔHf,ethane + .05ΔHf,N2 - ΔHf, CO2 - ΔHf,H2O(g)
The N2 does not react.

I don't follow what your friend did, but in part (a) you should have taken 0.85 times the heat of reaction 1 plus 0.10 times the heat of reaction 2. There is no change in the enthalpy of the N2 at 25C.

In part (b), I really like the way you did it. That's what I would have done.

Chet
 
Yes, I did that. Once you get the answer to the heat of combustion to part (a), you just add up the sensible heat changes. However, I actually changed my basis from 1 mol in part (a) to 100 moles in part (b), so I went back, changed the basis to 1 mol, and multiplied all the values I got for the sensible heats of the species by the number of moles coming out of the reactor.

It gives the right answer, but I was a little shaky why I had to do that. It seems like it has to do with the wording based on the fact that its per mole of fuel.
 
Maylis said:
Yes, I did that. Once you get the answer to the heat of combustion to part (a), you just add up the sensible heat changes. However, I actually changed my basis from 1 mol in part (a) to 100 moles in part (b), so I went back, changed the basis to 1 mol, and multiplied all the values I got for the sensible heats of the species by the number of moles coming out of the reactor.

It gives the right answer, but I was a little shaky why I had to do that. It seems like it has to do with the wording based on the fact that its per mole of fuel.
Yes. That's correct.

Chet
 
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